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A sequence $y=(y_n)_n$ is said to be a statistically convergent to $\lambda$ if for any $\epsilon>0$ the set $\{n\in\mathbb N: |y_n-\lambda|\geq\epsilon\}$ has natural density $0$.

Consider the following sequence :

$$x=(~\underbrace{0,0,\dots,0}_{100\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10\text{ copies}}~,~ \underbrace{0,0,\dots,0}_{100^2\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10^2\text{ copies}}~,~ \underbrace{0,0,\dots,0}_{100^3\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10^3\text{ copies}}~,~ \dots)$$

This sequence is not almost convergent because there exists no $\Omega\in \mathbb R$ such that $\lim\limits_{k\to \infty}\frac{x_{n+1}+x_{n+2}+\dots+x_{n+k}}{k}=\Omega$, for each $n\in \mathbb N$.

My Question : Is the sequence $x$ statistically convergent to $0$?

How can I prove or disprove that? How we deal here with natural density?

This sequence is from Proposition 1.1 in Miller, H. I.; Orhan, C., On almost convergent and statistically convergent subsequences, Acta Math. Hung. 93, No. 1-2, 135-151 (2001). ZBL0989.40002, MR1924673.

Martin Sleziak
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MAS
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    Can you define "statistically convergent to $0$" ? – Gabriel Romon Aug 30 '19 at 09:21
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    Also are you sure that the sequence is not almost convergent? It looks to me that for all $n\in \Bbb{N}$ we have that the limit equals $0$. We have that sum $\sum_{i=1}^{100^n}x_{i}\leq \sum_{i=1}^n 10^i\leq 10^{n+1}$ so we have that $\sum_{i=1}^{100^n}x_{i}/100^n\leq 10^{n+1}/100^n=10^{1-n}$ – kingW3 Aug 30 '19 at 09:32
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    @GabrielRomon I've edited my question by including the definition of statistical convergence. – MAS Aug 30 '19 at 09:36
  • @kingW3 In the following paper (see Proposition 1.1), I've seen this result. But I cannot understand the part of statistical convergence. – MAS Aug 30 '19 at 09:41
  • https://link.springer.com/article/10.1023/A:1013877718406 – MAS Aug 30 '19 at 10:10
  • @BijanDatta I have included the reference mentioned in your comment into the question, since it is a reasonable way to provide context. I will also mention that using citation helper (which is available when editing any post) you can get nicely formatted references relatively easily. – Martin Sleziak Sep 03 '19 at 06:30
  • @kingW3 I have tried to offer an argument why sequences of this form are not almost convergent in an answer to another question. The main thing here is that we have arbitrarily long segments consisting of zeroes and arbitrarily long segments consisting of ones. – Martin Sleziak Sep 03 '19 at 06:32

1 Answers1

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You have a sequence consisting of zeroes and ones. So it is statistically convergent to zero, if and only if the set $$A=\{n\in\mathbb N; x_n=1\}$$ has density zero, i.e., that $d(A)=0$.

Let $A(n)=|A\cap\{1,2,\dots,n\}$| denote the number of elements in $A$ which are less than or equal to $n$. Recall that the asymptotic density is defined as the limit of the fraction $A(n)/n$, taking limes superior/inferior we get the upper/lower asymptotic density.

I will also remind that in general when we have a set which has this type of "block structure", it suffices to check the value of $A(n)/n$ at the beginning end ends of the blocks to get the asymptotic density. See, for example, Proposition 2 in this answer.

In our case, the values of $A(n)/n$ at the ends of the blocks are $$\frac{10}{10+100},\frac{10+10^2}{10+100+10^2+100^2}, \frac{10+10^2+10^3}{10+100+10^2+100^2+10^3+100^3}.$$ Using Stolz-Cesaro theorem we can see that limit of the above sequence is equal to $$\lim_{n\to\infty} \frac{10^n}{10^n+100^n} = \lim_{n\to\infty} \frac{1}{1+10^n}=0.$$ This show that $\overline d(A)=0$ and, consequently, $d(A)=0$.

1I am using Stolz-Cesaro theorem in the "sum version", for comparing the two versions of this result see this answer. Of course, it would be easy to calculate the limit even without this theorem - we basically just need to compute the sums of two geometric series.

Martin Sleziak
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